Question

one mole of an ideal monoatomic gas is taken through the cycle shown in the figure.

the process A:B is a reversible isothermal expansion.

Given. P(A)=5atm V(A)=10litre

P(B)=1atm V(B)=50litre

P(C)=1atm V(C)=10liter

calculate the net work done by the gas

Answer 1

Solution.

$P_A=5atm=506625Pa;$

$V_A=10L=0.01m^3;$

$P_B=1atm=101325Pa;$

$V_B=50L=0.05m^3;$

$P_C=1 atm=101325Pa;$

$V_C=10L=0.01m^3;$

$P_AV_A=\nu RT_A\implies T_A=\dfrac{P_AV_A}{\nu R};$

$T_A=\dfrac{506625\sdot0.01}{1\sdot8.31}=609.66K;$

$P_CV_C=\nu RT_C\implies T_C=\dfrac{P_CV_C}{\nu R};$

$T_C=\dfrac{101325\sdot0.01}{1\sdot8.31}=121.93;$

$A=A_{AB}+A_{BC};$

$A_{AB}=\nu RTln\dfrac{V_2}{V_1};$

$A_{AB}=1\sdot8.31\sdot609.66\sdot ln\dfrac{0.05}{0.01}=8153.7J=8.1537kJ;$

$A_{BC}=P_B(V_C-V_B);$

$A_{BC}=101325\sdot(0.01-0.05)=-4053J=-4.053kJ;$

$A=8.1537-4.053=4.045kJ;$

Answer: $A=4.045kJ.$